Largest product of a subarray of size k
Given an array consisting of n positive integers, and an integer k. Find the largest product subarray of size k, i.e., find maximum produce of k contiguous elements in the array where k <= n.
Examples :
Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2} k = 6 Output: 4608 The subarray is {9, 8, 2, 4, 1, 8} Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2} k = 4 Output: 720 The subarray is {5, 9, 8, 2} Input: arr[] = {2, 5, 8, 1, 1, 3}; k = 3 Output: 80 The subarray is {2, 5, 8}
Method 1 (Simple : O(n*k))
A Naive approach would be to consider all the subarrays of size k one by one. Such a approach would require two loops hence the complexity would be O(n*k).
Method 2 (Efficient : O(n))
We can solve it in O(n) by using the fact that product of a subarray of size k can be computed in O(1) time if we have product of previous subarray available with us.
curr_product = (prev_product / arr[i-1]) * arr[i + k -1] prev_product : Product of subarray of size k beginning with arr[i-1] curr_product : Product of subarray of size k beginning with arr[i]
In this way, we can compute the maximum k size subarray product in only one traversal. Below is C++ implementation of the idea.
C++
// C++ program to find the maximum product of a subarray // of size k. #include <bits/stdc++.h> using namespace std; // This function returns maximum product of a subarray // of size k in given array, arr[0..n-1]. This function // assumes that k is smaller than or equal to n. int findMaxProduct( int arr[], int n, int k) { // Initialize the MaxProduct to 1, as all elements // in the array are positive int MaxProduct = 1; for ( int i=0; i<k; i++) MaxProduct *= arr[i]; int prev_product = MaxProduct; // Consider every product beginning with arr[i] // where i varies from 1 to n-k-1 for ( int i=1; i<=n-k; i++) { int curr_product = (prev_product/arr[i-1]) * arr[i+k-1]; MaxProduct = max(MaxProduct, curr_product); prev_product = curr_product; } // Return the maximum product found return MaxProduct; } // Driver code int main() { int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}; int k = 6; int n = sizeof (arr1)/ sizeof (arr1[0]); cout << findMaxProduct(arr1, n, k) << endl; k = 4; cout << findMaxProduct(arr1, n, k) << endl; int arr2[] = {2, 5, 8, 1, 1, 3}; k = 3; n = sizeof (arr2)/ sizeof (arr2[0]); cout << findMaxProduct(arr2, n, k); return 0; } |
Java
// Java program to find the maximum product of a subarray // of size k import java.io.*; import java.util.*; class GFG { // Function returns maximum product of a subarray // of size k in given array, arr[0..n-1]. This function // assumes that k is smaller than or equal to n. static int findMaxProduct( int arr[], int n, int k) { // Initialize the MaxProduct to 1, as all elements // in the array are positive int MaxProduct = 1 ; for ( int i= 0 ; i<k; i++) MaxProduct *= arr[i]; int prev_product = MaxProduct; // Consider every product beginning with arr[i] // where i varies from 1 to n-k-1 for ( int i= 1 ; i<=n-k; i++) { int curr_product = (prev_product/arr[i- 1 ]) * arr[i+k- 1 ]; MaxProduct = Math.max(MaxProduct, curr_product); prev_product = curr_product; } // Return the maximum product found return MaxProduct; } // driver program public static void main (String[] args) { int arr1[] = { 1 , 5 , 9 , 8 , 2 , 4 , 1 , 8 , 1 , 2 }; int k = 6 ; int n = arr1.length; System.out.println(findMaxProduct(arr1, n, k)); k = 4 ; System.out.println(findMaxProduct(arr1, n, k)); int arr2[] = { 2 , 5 , 8 , 1 , 1 , 3 }; k = 3 ; n = arr2.length; System.out.println(findMaxProduct(arr2, n, k)); } } // This code is contributed by Pramod Kumar |
Python3
# Python 3 program to find the maximum # product of a subarray of size k. # This function returns maximum product # of a subarray of size k in given array, # arr[0..n-1]. This function assumes # that k is smaller than or equal to n. def findMaxProduct(arr, n, k) : # Initialize the MaxProduct to 1, # as all elements in the array # are positive MaxProduct = 1 for i in range ( 0 , k) : MaxProduct = MaxProduct * arr[i] prev_product = MaxProduct # Consider every product beginning # with arr[i] where i varies from # 1 to n-k-1 for i in range ( 1 , n - k + 1 ) : curr_product = (prev_product / / arr[i - 1 ]) * arr[i + k - 1 ] MaxProduct = max (MaxProduct, curr_product) prev_product = curr_product # Return the maximum product found return MaxProduct # Driver code arr1 = [ 1 , 5 , 9 , 8 , 2 , 4 , 1 , 8 , 1 , 2 ] k = 6 n = len (arr1) print (findMaxProduct(arr1, n, k) ) k = 4 print (findMaxProduct(arr1, n, k)) arr2 = [ 2 , 5 , 8 , 1 , 1 , 3 ] k = 3 n = len (arr2) print (findMaxProduct(arr2, n, k)) # This code is contributed by Nikita Tiwari. |
C#
// C# program to find the maximum // product of a subarray of size k using System; class GFG { // Function returns maximum // product of a subarray of // size k in given array, // arr[0..n-1]. This function // assumes that k is smaller // than or equal to n. static int findMaxProduct( int []arr, int n, int k) { // Initialize the MaxProduct // to 1, as all elements // in the array are positive int MaxProduct = 1; for ( int i = 0; i < k; i++) MaxProduct *= arr[i]; int prev_product = MaxProduct; // Consider every product beginning // with arr[i] where i varies from // 1 to n-k-1 for ( int i = 1; i <= n - k; i++) { int curr_product = (prev_product / arr[i - 1]) * arr[i + k - 1]; MaxProduct = Math.Max(MaxProduct, curr_product); prev_product = curr_product; } // Return the maximum // product found return MaxProduct; } // Driver Code public static void Main () { int []arr1 = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}; int k = 6; int n = arr1.Length; Console.WriteLine(findMaxProduct(arr1, n, k)); k = 4; Console.WriteLine(findMaxProduct(arr1, n, k)); int []arr2 = {2, 5, 8, 1, 1, 3}; k = 3; n = arr2.Length; Console.WriteLine(findMaxProduct(arr2, n, k)); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP program to find the maximum // product of a subarray of size k. // This function returns maximum // product of a subarray of size // k in given array, arr[0..n-1]. // This function assumes that k // is smaller than or equal to n. function findMaxProduct( $arr , $n , $k ) { // Initialize the MaxProduct to // 1, as all elements // in the array are positive $MaxProduct = 1; for ( $i = 0; $i < $k ; $i ++) $MaxProduct *= $arr [ $i ]; $prev_product = $MaxProduct ; // Consider every product // beginning with arr[i] // where i varies from 1 // to n-k-1 for ( $i = 1; $i < $n - $k ; $i ++) { $curr_product = ( $prev_product / $arr [ $i - 1]) * $arr [ $i + $k - 1]; $MaxProduct = max( $MaxProduct , $curr_product ); $prev_product = $curr_product ; } // Return the maximum // product found return $MaxProduct ; } // Driver code $arr1 = array (1, 5, 9, 8, 2, 4, 1, 8, 1, 2); $k = 6; $n = count ( $arr1 ); echo findMaxProduct( $arr1 , $n , $k ), "\n" ; $k = 4; echo findMaxProduct( $arr1 , $n , $k ), "\n" ; $arr2 = array (2, 5, 8, 1, 1, 3); $k = 3; $n = count ( $arr2 ); echo findMaxProduct( $arr2 , $n , $k ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // JavaScript program to find the maximum // product of a subarray of size k // Function returns maximum // product of a subarray of // size k in given array, // arr[0..n-1]. This function // assumes that k is smaller // than or equal to n. function findMaxProduct(arr, n, k) { // Initialize the MaxProduct // to 1, as all elements // in the array are positive let MaxProduct = 1; for (let i = 0; i < k; i++) MaxProduct *= arr[i]; let prev_product = MaxProduct; // Consider every product beginning // with arr[i] where i varies from // 1 to n-k-1 for (let i = 1; i <= n - k; i++) { let curr_product = (prev_product / arr[i - 1]) * arr[i + k - 1]; MaxProduct = Math.max(MaxProduct, curr_product); prev_product = curr_product; } // Return the maximum // product found return MaxProduct; } let arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2]; let k = 6; let n = arr1.length; document.write(findMaxProduct(arr1, n, k) + "</br>" ); k = 4; document.write(findMaxProduct(arr1, n, k) + "</br>" ); let arr2 = [2, 5, 8, 1, 1, 3]; k = 3; n = arr2.length; document.write(findMaxProduct(arr2, n, k) + "</br>" ); </script> |
Output:
4608 720 80
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